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If the angle between the lines given by the equation
$x^{2}-3 x y+\lambda y^{2}+3 x-5 y+2=0, \lambda \geq 0$, is $\tan ^{-1}\left(\frac{1}{3}\right)$, then $\lambda=$
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$x^{2}-3 x y+\lambda y^{2}+3 x-5 y+2=0, \lambda \geq 0$, is $\tan ^{-1}\left(\frac{1}{3}\right)$, then $\lambda=$
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(D)
Given equation of pair of straight lines be $x^{2}-3 x y+\lambda y^{2}+3 x-5 y+2=0$.
Here $\mathrm{a}=1, \mathrm{~b}=\lambda, \mathrm{h}=\frac{-3}{2}$ and angle $\theta$ is given by $\tan ^{-1}\left(\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right)$
$\therefore \frac{1}{3}=\frac{2 \sqrt{\frac{9}{4}-\lambda}}{1+\lambda} \Rightarrow(1+\lambda)^{2}=\left(6 \sqrt{\frac{9}{4}-\lambda}\right)^{2}$
$\therefore 1+2 \lambda+\lambda^{2}=36\left(\frac{9}{4}-\lambda\right)=81-36 \lambda$
$\therefore \lambda^{2}+38 \lambda-80=0 \Rightarrow(\lambda+40)(\lambda-2)=0$
$\therefore \lambda=2,-40$
Given equation of pair of straight lines be $x^{2}-3 x y+\lambda y^{2}+3 x-5 y+2=0$.
Here $\mathrm{a}=1, \mathrm{~b}=\lambda, \mathrm{h}=\frac{-3}{2}$ and angle $\theta$ is given by $\tan ^{-1}\left(\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right)$
$\therefore \frac{1}{3}=\frac{2 \sqrt{\frac{9}{4}-\lambda}}{1+\lambda} \Rightarrow(1+\lambda)^{2}=\left(6 \sqrt{\frac{9}{4}-\lambda}\right)^{2}$
$\therefore 1+2 \lambda+\lambda^{2}=36\left(\frac{9}{4}-\lambda\right)=81-36 \lambda$
$\therefore \lambda^{2}+38 \lambda-80=0 \Rightarrow(\lambda+40)(\lambda-2)=0$
$\therefore \lambda=2,-40$
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